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  • Evaluate   <img alt="\mathrm{\lim _{x \rightarrow 4} \frac{a x-\sqrt{x}+b}{x-4}=\frac{3}{4}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarr

Evaluate   \mathrm{\lim _{x \rightarrow 4} \frac{a x-\sqrt{x}+b}{x-4}=\frac{3}{4}}

Option: 1

a = 1, b= -2


Option: 2

a = 3, b = -2


Option: 3

a = 5, b = -2


Option: 4

0


Answers (1)

best_answer

Substitute \mathrm{\sqrt{x}=y}, so that

\mathrm{\lim _{x \rightarrow 4} \frac{a x-\sqrt{x}+b}{x-4}=\lim _{y \rightarrow 2} \frac{a y^2-y+b}{y^2-4}=\frac{3}{4}}

We have 

\mathrm{\frac{a y^2-y+b}{y^2-4}=a-\frac{y-4 a-b}{(y-2)(y+2)}}

For the limit to exist, y = 2 must be a removable singularity , so y- 2 must divide \mathrm{y+4a+b}. We have

\mathrm{\frac{y-4 a-b}{y-2}=1-\frac{4 a+b-2}{y-2}}

and the remainder should be such that \mathrm{4a+b-2=0}. so,

\mathrm{\lim _{y \rightarrow 2} \frac{a y^2-y+b}{y^2-4}=\lim _{y \rightarrow 2}\left(a-\frac{1}{y+2}-\frac{4 a+b-2}{(y+2)(y-2)}\right)=a-\frac{1}{4}=\frac{3}{4}}

Then a =1 and b=-2

Posted by

Devendra Khairwa

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